Bounded Lipschitz norm ($\Vert f{\Vert }_{\mathrm{BL}}={\sup }_{x\ne y}|f(x)-f(y)|/d(x,y)+{\sup }_z|f(z)|$)

The bounded Lipschitz norm combines the Lipschitz constant and supremum norm of a bounded real-valued Lipschitz function on a metric space. Both summands satisfy the triangle inequality; the supremum term makes the sum definite even on constant functions. Its unit ball requires the sum of the two bounds to be at most one, a stronger requirement than bounding each separately by one.

On a separable metric space, a sequence uniformly bounded in bounded Lipschitz norm has a pointwise convergent subsequence. Enumerate a countable dense subset and use the Bolzano-Weierstrass theorem and diagonal subsequence argument to obtain convergence there. Uniform control of the Lipschitz constant then makes the subsequence Cauchy at every point. The pointwise closure of the bounded Lipschitz unit ball preserves the original norm bound after rescaling.

For every fixed $x\ne y$ and $z$, the sum $|f_i(x)-f_i(y)|/d(x,y)+|f_i(z)|$ is at most one. Pointwise convergence preserves this inequality. Taking the independent suprema over the pair $(x,y)$ and the point $z$ proves the assertion. Taking each bound separately would only give the weaker estimate two.